Question:$(3x+\frac{6}{y})+(\frac{x^2}{y}+\frac{3y}{x})\frac{dy}{dx}=0$
Solution:
From question, we know that:
$M=3x+\frac{6}{y}$
$N=(\frac{x^2}{y}+\frac{3y}{x})$
so we get: $My=6$, $Nx=\frac{2x}{y}-\frac{3y}{x^2}$
Observe that $My != Nx$, want to find $\mu$ to make it exact.
Since $R1=\frac{My-Nx}{M}$not a function of y only, and neither does $R2=\frac{My-Nx}{N}$ a function of x only,
use $R=\frac{Nx-My}{xM-Ny}$
$R=\frac{\frac{2x}{y}-\frac{3y}{x^2}+\frac{6}{y^2}}{3x^2+6\frac{x}{y}-x^2-\frac{3y^2}{x}}$
$R=\frac{2x^3y-3y^3+6x^2}{2x^3y+6x^2-3y^3}\frac{xy}{x^2y^2}$
$R=\frac{1}{xy}$
then $\mu=exp{(\int R(t) dt)}$ where t=xy
i.e $\mu(xy)=xy$
multiply $\mu$ to both sides and we get:
$(3x^2y+6x)+(x^3+3y^2)\frac{dy}{dx}=0$
Then the new $M=3x^2y+6x$, $N=x^3+3y^2$
$My=3x^2$, $Nx=3x^2$ so $My=Nx$, exact
Therefore there exists a function $\varphi(x,y)$ such that $\varphi x = M$, $\varphi y=N$
$\varphi x=3x^2y+6x$, integrate and we get $\varphi(x,y)=x^3y+3x^2+h(y)$
so take the derivative over y and we get $\varphi y =x^3+h'(y)$
since $N=x^3+3y^2$, so $h'(y)=3y^2$
so $h(y)=y^3$ from integration
therefore the general solution is $\varphi(x,y)= x^3y+3x^2+y^3$
Hence the solution of given equations are given implicitly by $x^3y+3x^2+y^3=C$
